Question of the Week - 37 (The nth term of a sequence...)

Solution

• The nth term of a sequence a1, a2, a3, … an is given by an = \(\frac{1}{n(n+2)}\)

• The sum of the first 20 terms of the sequence.

• Hence, a1 = \(\frac{1}{1(1+2)} =\frac{1}{(1×3)}\)
• Similarly, a2 = \(\frac{1}{(2(2+2))}=\frac{1}{(2×4)}\)

o a3 = \(\frac{1}{(3×5)}\)
o a4 =\(\frac{1}{(4×6)}\) and so on.
o a19 = \(\frac{1}{(19×21)}\)
o a20 = \(\frac{1}{(20×22)}\)

• Hence, a1 + a2 + a3 + … +a19 + a20 =\(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}\)

• S = \(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}\)

• 2S = 2(\(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}) 2\)
• 2S = \(\frac{2}{(1×3)}\) + \(\frac{2}{(2×4)}\) + \(\frac{2}{(3×5)}\) + \(\frac{2}{(4×6)}\) + ... + \(\frac{2}{(19×21)}\) + \(\frac{2}{(20×22)}\)

o If we observe carefully then,

• \(\frac{2}{(1×3)} = \frac{1}{1}- \frac{1}{3}\)
• \(\frac{2}{(2×4)}= \frac{1}{2}- \frac{1}{4}\)
• \(\frac{2}{(3×5)} = \frac{1}{3} - \frac{1}{5}\)
• \(\frac{2}{(4×6)} = \frac{1}{4} - \frac{1}{6}\)
• Similarly, \(\frac{2}{(19×21)} = \frac{1}{19}- \frac{1}{21}\) and \(\frac{2}{(20×22)} = \frac{1}{20} - \frac{1}{22}\)

o Now, we can observe that odd or even numbered terms are beginning with the same number that the preceding number is ending with and their sign is opposite.
o So, they will get cancelled when added.

• For example: \(\frac{2}{(1×3)} = \frac{1}{1} - \frac{1}{3} and \frac{2}{(3×5)} = \frac{1}{3} - \frac{1}{5}\) (odd numbered terms)
o 1/3 will get cancelled when these two terms are added.

• And, we will get 1- 1/5 i.e. First number of first term - last number of last term.
o Similarly, if we add the next odd term then 1/3 will get cancelled.
o Thus, if we add all the odd terms till a20 then:

• Sum = First number of first odd term - last number of last odd term.


• \(\frac{2}{(2×4)} = \frac{1}{2} - \frac{1}{4}\) and \(\frac{2}{(4×6)} = \frac{1}{4} - \frac{1}{6}\) (Even numbered terms)

o As we did with odd terms, the sum of all the even terms till a20 = First number of first even term - last number of last even term.

• Or, 2S = Sum of all odd terms till a20 + Sum of all even terms till a20

o = (2a1 + 2a3 + ... + 2a19) + (2a2 + 2a4 + ... + 2a20)
o = \([ (\frac{1}{1} - \frac{1}{3})+ ... +(\frac{1}{19} - \frac{1}{21}) ] + [ (\frac{1}{2} - \frac{1}{4})+ … +(\frac{1}{20} - \frac{1}{22}) ]\)
o = \((\frac{1}{1} - \frac{1}{21}) + (\frac{1}{2} - \frac{1}{22})\)
o 2S = \(\frac{20}{21} + \frac{10}{22}\)
o S = \(\frac{10}{21}+ \frac{5}{22}\)
o S = \(\frac{325}{462}\)